IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

tka

Q. The resultant pH when 200 ml of an aqueous solution of HCl (pH 2) and 300 ml of NaOH (pH 12) is:

a. 11.3

b. 7.0

c. 2.0

d. 13.0

It's obviously basic, but how do we calculate?

catch_arnnie

i guess the following should be the solution...

[H+](means H+ ion concentration) =10^-2 M

[OH-]=10^-2 M

millimoles of H+ = 10^-2 * 200 = 2

millimoles of OH- = 10^-2 * 300 = 3

NaOH + HCl ----> NaCl + H20

3 2 0 0 ----->initially

1 0 2 2 --->after neutralisation

=> 1 millimole of OH- is present in 500 ml water

=>[OH-] = 1/500 = 2 * 10^-3

=>pOH = 2.69

=>pH = 11.3 approx.

i think option (a) is correct..

tka

hey man, thats right, thanks a lot